Calculate the magnetic moment of $K_4[Mn(CN)_6]$ complex.

1.73 BM

The correct answer is Option (3) → 1.73 BM

To calculate the magnetic moment of $K_4​[Mn(CN)_6​]$, we follow a systematic approach using the oxidation state and ligand field strength.

1. Determine the Oxidation State of Manganese

The complex $K_4​[Mn(CN)_6​]$ is electrically neutral.

• Potassium (K): Each K has a charge of +1. Total for 4 atoms = +4.
• Cyanide (CN): Each CN is a ligand with a charge of −1. Total for 6 ligands = −6.
• Let x be the oxidation state of Mn:

$4(+1)+x+6(−1)=0$

$4+x−6=0⟹x=+2$

The oxidation state of Manganese is +2.

2. Determine the Electronic Configuration

• Neutral Mn (Z=25): $[Ar]3d^54s^2$
• $Mn^{2+}$ ion: Losing two electrons from the 4s orbital gives $[Ar]3d^5$.

3. Consider the Ligand Field ($CN^−$)

Cyanide ($CN^−$) is a strong field ligand. According to Crystal Field Theory, in an octahedral complex, strong field ligands cause the pairing of electrons in the lower energy $t_{2g}$​ orbitals.

• Configuration: The 5 electrons in the 3d subshell will pair up as much as possible in the $t_{2g}$​ level.
• Distribution: $t_{2g}5​_eg^0$​
• Unpaired Electrons (n): Out of 5 electrons, 4 will pair up (2 pairs), leaving n=1 unpaired electron.

4. Calculate Magnetic Moment (μ)

Using the spin-only magnetic moment formula:

$μ=\sqrt{n(n+2)}​ BM$

$μ=\sqrt{1(1+2)}​=\sqrt{3}​≈1.732 BM$